Calculus II · Unit 4A · lesson

Explicit and Recursive Sequences

Translate between explicit and recursive sequence rules, identify initial conditions, and analyze how step-by-step definitions generate long-term behavior.

Core textbook pathFollow the sequence or choose a destination from the unit map.Open the Unit 4A map →

Section overview

Sequences and their limits

What this section is building

Translate between explicit and recursive sequence rules, identify initial conditions, and analyze how step-by-step definitions generate long-term behavior.

Notice

A sequence converges when every sufficiently late term remains arbitrarily close to one finite target.

Decide

Use algebraic limits when a formula is explicit; use bounds and monotonicity when a recurrence hides the formula.

Avoid

Reading a finite plot as proof or solving a recurrence's fixed-point equation before proving convergence.

Concept

Learning objectives

convert between explicit and recursive descriptions and recognize arithmetic and geometric patterns.

Explicit and Recursive Sequences

Explanation

Two legitimate ways to describe the same list

An explicit formula calculates $a_n$ directly from $n$ . A recursive formula calculates a new term from earlier terms and therefore must include enough starting information to begin the process. Explicit descriptions are convenient for limits and distant terms; recursive descriptions often reflect how a process actually evolves one step at a time.

Arithmetic sequences add a constant difference, while geometric sequences multiply by a constant ratio. Those are not merely vocabulary categories. They predict closed forms, growth rates, and later the behavior of geometric series. When translating, check both the initial term and the update rule; matching only one of them produces a sequence that looks plausible for a few steps and is still wrong.

Bridge

Description versus process

An explicit rule answers, "What is the $n$ th term?" A recurrence answers, "How does the process move from one stage to the next?" The two descriptions define the same sequence only when the starting data and the update rule agree.

Explicit formulas are convenient for distant terms and limits. Recursive formulas often mirror how an account, population, algorithm, or physical system actually evolves. Writing several terms before translating exposes the starting index and prevents a plausible-looking formula from describing the wrong list.

**One sequence, two descriptions**
The same arithmetic sequence can be generated step by step or evaluated directly.
One sequence, two descriptions. Side-by-side recurrence chain and direct formula.

Concept

Two standard families

If $a_1=A$ , then

\text{arithmetic: }a_n=A+(n-1)d,\qquad \text{geometric: }a_n=Ar^{n-1}.

Recursively these become $a_n=a_{n-1}+d$ and $a_n=ra_{n-1}$ .

Guided walkthrough

Convert a recurrence to an explicit rule

Suppose $a_1=3$ and $a_n=a_{n-1}+4$ . The list is $3,7,11,15,\ldots$ , an arithmetic sequence with difference $4$ . Therefore

a_n=3+4(n-1)=4n-1.

Checking $n=1$ gives $3$ , and increasing $n$ by one increases the formula by $4$ .

Worked example

A recurrence that is not arithmetic or geometric

Let $a_1=1$ and $a_{n+1}=(a_n+3)/2$ . The first terms are

1,\ 2,\ \frac52,\ \frac{11}{4},\ \frac{23}{8},\ldots

There is no constant difference or ratio. If the sequence converges to $L$ , then the shifted sequence has the same limit, so

L=\frac{L+3}{2},

which gives $L=3$ . This identifies the only possible limit; it does not yet prove convergence. That proof will require monotonicity and boundedness.

Common mistake

A recurrence needs enough starting data

The rule $a_n=a_{n-1}+a_{n-2}$ requires two initial values. One value does not determine the next term and therefore does not determine a unique sequence.

Interactive checku4a-explicit_and_recursive_sequences-01

If $a_1=3$ and $a_n=a_{n-1}+4$ , find $a_5$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Apply the update four times or use the explicit formula.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Write a recursion for $5,15,45,135,\ldots$ .

Exercise

Find an explicit formula for $12,9,6,3,\ldots$ .

Exercise

Explain why the recurrence $a_n=a_{n-1}+a_{n-2}$ needs two initial values.

Exercise

Model annual account growth of 6 percent with a recursive sequence.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Explicit and Recursive Sequences

Explanation

Two legitimate ways to describe the same list

Bridge

Description versus process

Read this graph as text

One sequence, two descriptions. The same arithmetic sequence can be generated step by step or evaluated directly. Side-by-side recurrence chain and direct formula. Preserve the misconception control described in the lesson and storyboard.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in one sequence, two descriptions; color is never the only cue.

Why it matters: Side-by-side recurrence chain and direct formula.

One sequence, two descriptions

The same arithmetic sequence can be generated step by step or evaluated directly.

One sequence, two descriptions. Side-by-side recurrence chain and direct formula.

Concept

Two standard families

If $a_1=A$ , then

\text{arithmetic: }a_n=A+(n-1)d,\qquad \text{geometric: }a_n=Ar^{n-1}.

Recursively these become $a_n=a_{n-1}+d$ and $a_n=ra_{n-1}$ .

Guided walkthrough

Convert a recurrence to an explicit rule

Suppose $a_1=3$ and $a_n=a_{n-1}+4$ . The list is $3,7,11,15,\ldots$ , an arithmetic sequence with difference $4$ . Therefore

a_n=3+4(n-1)=4n-1.

Checking $n=1$ gives $3$ , and increasing $n$ by one increases the formula by $4$ .

Worked example

A recurrence that is not arithmetic or geometric

Let $a_1=1$ and $a_{n+1}=(a_n+3)/2$ . The first terms are

1,\ 2,\ \frac52,\ \frac{11}{4},\ \frac{23}{8},\ldots

There is no constant difference or ratio. If the sequence converges to $L$ , then the shifted sequence has the same limit, so

L=\frac{L+3}{2},

which gives $L=3$ . This identifies the only possible limit; it does not yet prove convergence. That proof will require monotonicity and boundedness.

Common mistake

A recurrence needs enough starting data

The rule $a_n=a_{n-1}+a_{n-2}$ requires two initial values. One value does not determine the next term and therefore does not determine a unique sequence.

Interactive checku4a-explicit_and_recursive_sequences-01

If $a_1=3$ and $a_n=a_{n-1}+4$ , find $a_5$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Apply the update four times or use the explicit formula.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Write a recursion for $5,15,45,135,\ldots$ .

Exercise

Find an explicit formula for $12,9,6,3,\ldots$ .

Exercise

Explain why the recurrence $a_n=a_{n-1}+a_{n-2}$ needs two initial values.

Exercise

Model annual account growth of 6 percent with a recursive sequence.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Explicit and Recursive Sequences

Related resources

Original instruction with traceable references.

Explicit and Recursive Sequences

Related resources

Original instruction with traceable references.